\newproblem{lay:7_4_15}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.4.15}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Suppose the factorization below is an SVD of a matrix $A$, with the entries $U$ and $V$ rounded to two decimal places.
	\begin{center}
		$A=\begin{pmatrix}0.40 & -0.78 & 0.47 \\ 0.37 & -0.33 & -0.87\\ -0.84 & -0.52 & -0.16\end{pmatrix}
		   \begin{pmatrix}7.10 & 0 & 0 \\ 0 & 3.10 & 0 \\ 0 & 0 & 0\end{pmatrix}
			 \begin{pmatrix}0.30 & -0.51 & -0.81 \\ 0.76 & 0.64 & -0.12 \\ 0.58 & -0.58 & 0.58\end{pmatrix}$
	\end{center}
	\begin{enumerate}[a.]
		\item What is the rank of $A$?
		\item Use the decomposition of $A$, with no calculations, to write a basis for $\mathrm{Col}\{A\}$ and a basis for $\mathrm{Nul}\{A\}$.
		      [\textit{Hint}: First write the columns of $V$.]
	\end{enumerate}
}{
   % Solution
	The factorization above is of the form $A=U\Sigma V^T$. $A$ is a $3\times 3$ matrix ($n=3$).
	\begin{enumerate}[a.]
		\item Since $A$ has only two non-zero singular values, its rank is 2.
		\item $V$ is $\begin{pmatrix}0.30 & 0.76 & 0.58 \\ -0.51 & 0.64 & -0.58 \\ -0.81 & -0.12 & 0.58\end{pmatrix}$. Since $A$ is of rank $2$, the first two columns
		      of $U$ provide a basis for $\mathrm{Col}\{A\}$
					\begin{center}
						$\mathrm{Basis}\{\mathrm{Col}\{A\}\}=\{(0.40,0.37,-0.84),(-0.78,-0.33,-0.52)\}$
					\end{center}
					Also, the last column ($n-r=3-2=1$) of $V$ provides a basis for $\mathrm{Nul}\{A\}$
					\begin{center}
						$\mathrm{Basis}\{\mathrm{Nul}\{A\}\}=\{(0.58,-0.58,0.58)\}$
					\end{center}
	\end{enumerate}
}
\useproblem{lay:7_4_15}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

